On the Nonoscillation of Second-Order Neutral Delay Differential Equation with Forcing Term

Abstract

This paper is concerned with nonoscillation of second-order neutral delay differential equation with forcing term. By using contraction mapping principle, some sufficient conditions for the existence of nonoscillatory solution are established. The criteria obtained in this paper complement and extend several known results in the literature. Some examples illustrating our main results are given.

1. Introduction

During the last two decades, there has been much research activity concerning the oscillation and nonoscillation of solutions of neutral type delay differential equations (see [1–9]). Investigation of such equations or systems, besides of their theoretical interest, have some importance in modelling of the networks containing lossless transmission lines, in the study of vibrating masses attached to an elastic bar and also in population dynamics, and so forth (see [1, 2, 8, 10] and the references cited therein).

In this paper, we consider the second-order neutral delay differential equation with forcing term of the form𝑥(𝑡)+𝑃(𝑡)𝑥(𝑡−𝜏)^′′+𝑄₁(𝑡)𝑥𝑡−𝜎₁−𝑄₂(𝑡)𝑥𝑡−𝜎₂=ℎ(𝑡),𝑡≥𝑡₀, (1.1)where𝜏>0,𝜎₁≥0,𝜎₂≥0,𝑃,𝑄₁,𝑄₂,ℎ∈𝐶𝑡₀,∞,ℝ. (1.2)

Let 𝜑∈𝐶([𝑡₀−𝜎,𝑡₀),ℝ), where 𝜎=max{𝜏,𝜎₁,𝜎₂}, be a given function and let 𝑥₀ be a given constant. By the method of steps (see [2]), it is easy to know that (1.1) has a unique solution 𝑥∈𝐶([𝑡₀−𝜎,∞),ℝ) in the sense that 𝑥(𝑡)+𝑃(𝑡)𝑥(𝑡−𝜏) is twice continuously differential for 𝑡≥𝑡₀, 𝑥(𝑡) satisfies (1.1) and𝑥(𝑠)=𝜑(𝑠)for𝑠∈[𝑡₀−𝜎,𝑡₀],𝑥(𝑡)+𝑃(𝑡)𝑥(𝑡−𝜏))^′_𝑡=𝑡0=𝑥₀. (1.3)

As is customary, a solution of (1.1) is called oscillatory if it has arbitrarily large zeros, and otherwise it is nonoscillatory. Equation (1.1) is oscillatory if all its solutions are oscillatory.

When 𝑃(𝑡)=𝑝 and the forcing term ℎ(𝑡)≡0, (1.1) reduces to𝑥(𝑡)+𝑝𝑥(𝑡−𝜏)^′′+𝑄₁(𝑡)𝑥𝑡−𝜎₁−𝑄₂(𝑡)𝑥𝑡−𝜎₂=0,𝑡≥𝑡₀, (1.4) where 𝑝∈ℝ(𝑝≠±1) and ∫^∞𝑡𝑄_𝑖(𝑡)𝑑𝑡<∞,𝑖=1,2. The first global result of (1.4) (with respect to 𝑝), which is a sufficient condition for the existence of a nonoscillatory solution for all values of 𝑝≠±1, have been examined by Kulenović and Hadžiomerspahić [4].

Recently, Parhi and Rath [7] studied oscillation behaviors for forced first-order neutral differential equations as follows 𝑥(𝑡)−𝑃(𝑡)𝑥(𝑡−𝜏)^′+𝑄(𝑡)𝐺𝑥(𝑡−𝜎)=ℎ(𝑡),𝑡≥𝑡₀. (1.5) Necessary and sufficient conditions are obtained in various ranges for 𝑃(𝑡)≠±1 so that every solution of (1.5) is oscillatory or tends to zero or to ±∞ as 𝑡→∞.

Motivated by the idea of [4, 7], in present paper we establish sufficient conditions for existence of a nonoscillatory solution to (1.1) depending on various ranges of 𝑃(𝑡)≠±1. Hereinafter, we assume that the following conditions hold,

(H1)𝑄_𝑖≥0, and ∫^∞𝑡𝑄_𝑖(𝑡)𝑑𝑡<∞,𝑖=1,2.(H2)There exists a function 𝐻(𝑡)∈𝐶²([𝑡₀,∞),ℝ) such that 𝐻^′′(𝑡)=ℎ(𝑡) and lim_𝑡→∞𝐻(𝑡)=𝑀∈ℝ.

2. Main results

Theorem 2.1. Suppose that conditions (𝐻1) and (𝐻2) hold. If 𝑃(𝑡) is in one of the following ranges: (i)0<𝑃(𝑡)≤𝑝₁<1,(ii)1<𝑝₂≤𝑃(𝑡)≤𝑝₁,(iii)−1<−𝑝₂≤𝑃(𝑡)<0,(iv)−𝑝₂≤𝑃(𝑡)≤−𝑝₁<−1, (2.1) then (1.1) has a nonoscillatory solution.

Proof. The proof of this theorem will be divided into four cases in terms of the four different ranges of 𝑃(𝑡).
Case (i) (0<𝑃(𝑡)≤𝑝₁<1). Choose a 𝑡₁>𝑡₀+𝜎 sufficiently large such that ^∞_𝑡1𝑠𝑄₁(𝑠)+𝑄₂3(𝑠)𝑑𝑠<1−𝑝₁4_∞𝑡1,𝑠𝑄₁𝑝(𝑠)𝑑𝑠≤¹+𝑁₁−1𝑁_1∞𝑡1,𝑠𝑄₂(𝑠)𝑑𝑠≤1−𝑝₁1+2𝑁₁−2𝑀₁2𝑁₁,𝐻(𝑡)−𝑀≤1−𝑝₁4, (2.2)(2.3)(2.4)(2.5) where 𝑀₁ and 𝑁₁ are positive constants such that1−𝑁₁<𝑝₁<1−2𝑀₁1+2𝑁₁. (2.6)
Let 𝑋 be the set of all continuous and bounded functions on [𝑡₀,∞) with the sup norm. Set𝐴=𝑥∈𝑋∶𝑀₁≤𝑥(𝑡)≤𝑁₁,𝑡≥𝑡₀. (2.7)Define a mapping 𝑇∶𝐴→𝑋 as follows:3(𝑇𝑥)(𝑡)=1−𝑝₁4−𝑃(𝑡)𝑥(𝑡−𝜏)+𝑡^∞_𝑡𝑄₁(𝑠)𝑥𝑠−𝜎₁−𝑄₂(𝑠)𝑥𝑠−𝜎₂+𝑑𝑠_𝑡𝑡1𝑠𝑄₁(𝑠)𝑥𝑠−𝜎₁−𝑄₂(𝑠)𝑥𝑠−𝜎₂𝑑𝑠+𝐻(𝑡)−𝑀,𝑡≥𝑡₁(𝑇𝑥)𝑡₁,𝑡₀≤𝑡≤𝑡₁. (2.8)
Clearly, 𝑇𝑥 is continuous. For every 𝑥∈𝐴 and 𝑡≥𝑡₁, using (2.3) and (2.5), we get (𝑇𝑥)(𝑡)≤1−𝑝₁+𝑁₁^∞_𝑡1𝑠𝑄₁(𝑠)𝑑𝑠≤𝑁₁. (2.9) Furthermore, from (2.4) and (2.5), we have(𝑇𝑥)(𝑡)≥1−𝑝₁2−𝑝₁𝑁₁−𝑁₁^∞_𝑡1𝑠𝑄₂(𝑠)𝑑𝑠≥𝑀₁. (2.10)Thus we prove that 𝑇𝐴⊂𝐴. To apply the contraction principle, we need to prove that 𝑇 is a contraction mapping on 𝐴 since 𝐴 is a bounded, closed, and convex subset of 𝑋.
Now, for 𝑥₁,𝑥₂∈𝐴 and 𝑡≥𝑡₁ we have𝑇𝑥₁(𝑡)−𝑇𝑥₂(𝑡)≤𝑥₁−𝑥₂𝑝₁+^∞_𝑡1𝑠𝑄₁(𝑠)+𝑄₂(𝑠)𝑑𝑠=𝑞₁𝑥₁−𝑥₂, (2.11)where we used sup norm. From (2.2), we obtain 𝑞₁<1, which completes the proof of Case (i).
Example 2.2. Consider the second-order neutral delay differential equation𝑥(𝑡)+𝑒^−𝑡𝑥(𝑡−1)^′′+𝑒^−𝑡−1𝑥(𝑡−1)−4𝑒^−𝑡𝑥(𝑡−1)=ℎ(𝑡),𝑡≥1, (2.12)where 𝑃(𝑡)=𝑒^−𝑡,𝑄₁(𝑡)=𝑒^−𝑡−1,𝑄₂(𝑡)=4𝑒^−𝑡,ℎ(𝑡)=𝑒^−𝑡+𝑒^−2𝑡 such that 0<𝑃(𝑡)≤𝑒⁻¹<1. Since 𝐻(𝑡)=𝑒^−𝑡+1/4𝑒^−2𝑡→0 as 𝑡→∞, ∫^∞_𝑡1𝑠𝑄₁(𝑠)𝑑𝑠=2𝑒^−𝑡, and ∫^∞_𝑡1𝑠𝑄₂(𝑠)𝑑𝑠=8𝑒^−𝑡, then the sufficient conditions—in Case (i) of Theorem 2.1—are satisfied. Therefore, the equation has a positive solution. In fact 𝑦=𝑒^−𝑡 is a positive solution of this equation.
Case (ii) (1<𝑝₂≤𝑃(𝑡)≤𝑝₁). Choose a 𝑡₁>𝑡₀+𝜎 sufficiently large such that ^∞_𝑡1𝑠𝑄₁(𝑠)+𝑄₂3𝑝(𝑠)𝑑𝑠<²−14_∞𝑡1,𝑠𝑄₁(𝑠)𝑑𝑠≤1−𝑝₂1−𝑁₂𝑁_2∞𝑡1,𝑠𝑄₂𝑝(𝑠)𝑑𝑠≤²𝑝₂−1−2𝑝₁𝑁₂+𝑝₂𝑀₂2𝑝₁𝑁₂,≤𝑝𝐻(𝑡)−𝑀²−14, (2.13)(2.14)(2.15)(2.16)where 𝑀₂ and 𝑁₂ are positive constants such that11−𝑝₂<𝑁₂<𝑝₂𝑝₂−1−2𝑝₁𝑀₂2𝑝₁. (2.17)
Let 𝑋 be the set as in Case (i). Set 𝐴=𝑥∈𝑋∶𝑀₂≤𝑥(𝑡)≤𝑁₂,𝑡≥𝑡₀. (2.18)Define a mapping 𝑇∶𝐴→𝑋 as follows:=𝑝(𝑇𝑥)(𝑡)₂−1−4𝑃(𝑡+𝜏)𝑥(𝑡+𝜏)+𝑃(𝑡+𝜏)𝑡+𝜏𝑃(𝑡+𝜏)^∞_𝑡+𝜏𝑄₁(𝑠)𝑥𝑠−𝜎₁−𝑄₂(𝑠)𝑥𝑠−𝜎₂+1𝑑𝑠𝑃(𝑡+𝜏)_𝑡^𝑡+𝜏₁𝑠𝑄₁(𝑠)𝑥𝑠−𝜎₁−𝑄₂(𝑠)𝑥𝑠−𝜎₂𝑑𝑠+𝐻(𝑡+𝜏)−𝑀𝑃(𝑡+𝜏),𝑡≥𝑡₁(𝑇𝑥)𝑡₁,𝑡₀≤𝑡≤𝑡₁. (2.19)
Clearly, 𝑇𝑥 is continuous. For every 𝑥∈𝐴 and 𝑡≥𝑡₁, using (2.14) and (2.16), we get1(𝑇𝑥)(𝑡)≤1−𝑝₂+𝑁₂𝑝₂^∞_𝑡1𝑠𝑄₁(𝑠)𝑑𝑠≤𝑁₂. (2.20)Furthermore, from (2.15) and (2.16) we have𝑝(𝑇𝑥)(𝑡)≥²−12𝑝₁−𝑁₂𝑝₂−𝑁₂𝑝₂^∞_𝑡1𝑠𝑄₂(𝑠)𝑑𝑠≥𝑀₂. (2.21)Thus we prove that 𝑇𝐴⊂𝐴. To apply the contraction principle, we need to prove that 𝑇 is a contraction mapping on 𝐴 since 𝐴 is a bounded, closed, and convex subset of 𝑋.
Now, for 𝑥₁,𝑥₂∈𝐴 and 𝑡≥𝑡₁, we have𝑇𝑥₁(𝑡)−𝑇𝑥₂≤1(𝑡)𝑝₂𝑥₁−𝑥₂1+^∞_𝑡1𝑠𝑄₁(𝑠)+𝑄₂(𝑠)𝑑𝑠=𝑞₂𝑥₁−𝑥₂, (2.22)where we used sup norm. From (2.13), we obtain 𝑞₂<1, which completes the proof of Case (ii).
Example 2.3. Consider the second-order neutral delay differential equation𝑥(𝑡)+2+𝑒^−𝑡𝑥(𝑡−1)^′′+𝑒^−𝑡−1𝑥(𝑡−1)−4𝑒^−𝑡𝑥(𝑡−1))=ℎ(𝑡),𝑡≥1, (2.23)where 𝑃(𝑡)=2+𝑒^−𝑡,𝑄₁(𝑡)=𝑒^−𝑡−1,𝑄₂(𝑡)=4𝑒^−𝑡,ℎ(𝑡)=𝑒^−𝑡+2𝑒^−𝑡+1+𝑒^−2𝑡 such that 𝑃(𝑡)≥2+𝑒⁻¹>1. Since 𝐻(𝑡)=𝑒^−𝑡+1/4𝑒^−2𝑡+2𝑒^−𝑡+1→0 as 𝑡→∞, ∫^∞_𝑡1𝑠𝑄₁(𝑠)𝑑𝑠=2𝑒^−𝑡, and ∫^∞_𝑡1𝑠𝑄₂(𝑠)𝑑𝑠=8𝑒^−𝑡, then the sufficient conditions—in Case (ii) of Theorem 2.1—are satisfied. Therefore, the equation has a positive solution. In fact 𝑦=𝑒^−𝑡 is a positive solution of this equation.
Case (iii) (−1<−𝑝₂≤𝑃(𝑡)<0). Choose a 𝑡₁>𝑡₀+𝜎 sufficiently large such that ^∞_𝑡1𝑠𝑄₁(𝑠)+𝑄₂3(𝑠)𝑑𝑠<1−𝑝₂4_∞𝑡1,𝑠𝑄₁(𝑠)𝑑𝑠≤1−𝑝₂𝑁₃−1𝑁_3∞𝑡1,𝑠𝑄₂(𝑠)𝑑𝑠≤1−𝑝₂−2𝑀₃2𝑁₃,𝐻(𝑡)−𝑀≤1−𝑝₂4, (2.24)(2.25)(2.26)(2.27)where 𝑀₃ and 𝑁₃ are positive constants such that2𝑀₃+𝑝₂<1<𝑁₃. (2.28)
Let 𝑋 be the set as in Case (i). Set 𝐴=𝑥∈𝑋∶𝑀₃≤𝑥(𝑡)≤𝑁₃,𝑡≥𝑡₀. (2.29)Define a mapping 𝑇∶𝐴→𝑋 as follows:3(𝑇𝑥)(𝑡)=1−𝑝₂4−𝑃(𝑡)𝑥(𝑡−𝜏)+𝑡^∞_𝑡1𝑄₁(𝑠)𝑥𝑠−𝜎₁−𝑄₂(𝑠)𝑥𝑠−𝜎₂+𝑑𝑠_𝑡𝑡1𝑠𝑄₁(𝑠)𝑥𝑠−𝜎₁−𝑄₂(𝑠)𝑥𝑠−𝜎₂𝑑𝑠+𝐻(𝑡)−𝑀,𝑡≥𝑡₁(𝑇𝑥)𝑡₁,𝑡₀≤𝑡≤𝑡₁. (2.30)
Clearly, 𝑇𝑥 is continuous. For every 𝑥∈𝐴 and 𝑡≥𝑡₁, using (2.25) and (2.27), we get(𝑇𝑥)(𝑡)≤1−𝑝₂+𝑝₂𝑁₃+𝑁₃^∞_𝑡1𝑠𝑄₁(𝑠)𝑑𝑠≤𝑁₃. (2.31)Furthermore, from (2.26) and (2.27), we have(𝑇𝑥)(𝑡)≥1−𝑝₂2−𝑁₃^∞_𝑡1𝑠𝑄₂(𝑠)𝑑𝑠≥𝑀₃. (2.32)Thus we prove that 𝑇𝐴⊂𝐴. To apply the contraction principle, we need to prove that 𝑇 is a contraction mapping on 𝐴 since 𝐴 is a bounded, closed, and convex subset of 𝑋.
Now, for 𝑥₁,𝑥₂∈𝐴 and 𝑡≥𝑡₁, we have𝑇𝑥₁(𝑡)−𝑇𝑥₂(𝑡)≤𝑥₁−𝑥₂𝑝₂+^∞_𝑡1𝑠𝑄₁(𝑠)+𝑄₂(𝑠)𝑑𝑠=𝑞₃𝑥₁−𝑥₂, (2.33)where we used sup norm. From (2.24), we obtain 𝑞₃<1, which completes the proof of Case (iii).
Example 2.4. Consider the second-order neutral delay differential equation 𝑥(𝑡)−𝑒^−𝑡𝑥(𝑡−1)^′′+𝑒^−𝑡−1𝑥(𝑡−1)−4𝑒^−𝑡𝑥(𝑡−1)=ℎ(𝑡),𝑡≥1, (2.34)where ℎ(𝑡)=𝑒^−𝑡+𝑒^−2𝑡−8𝑒^−2𝑡+1. This equation has a nonoscillatory solution 𝑦=𝑒^−𝑡 since the sufficient conditions—in Case (iii) of Theorem 2.1—are satisfied.
Case (iv) (−𝑝₂≤𝑃(𝑡)≤−𝑝₁<−1). Choose a 𝑡₁>𝑡₀+𝜎 sufficiently large such that^∞_𝑡1𝑠𝑄₁(𝑠)+𝑄₂3𝑝(𝑠)𝑑𝑠<¹−14_∞𝑡1,𝑠𝑄₁𝑁(𝑠)𝑑𝑠≤⁴𝑝₁−1−𝑝₁𝑁_4∞𝑡1,𝑠𝑄₂𝑝(𝑠)𝑑𝑠≤²−𝑝₁1+𝑀₄𝑝₂−1𝑝₂𝑁₄,𝐻(𝑡)−𝑀≤𝑝₁−1, (2.35)(2.36)(2.37)(2.38)where 𝑀₄ and 𝑁₄ are positive constants such that𝑁₄>𝑝₁𝑝₁−1,𝑀₄<𝑝₂−𝑝₁𝑝₂−1𝑝₁𝑝₂−1. (2.39)
Let 𝑋 be the set as in Case (i). Set𝐴=𝑥∈𝑋∶𝑀₄≤𝑥(𝑡)≤𝑁₄,𝑡≥𝑡₀. (2.40)Define a mapping 𝑇∶𝐴→𝑋 as follows=−1(𝑇𝑥)(𝑡)−𝑃(𝑡+𝜏)𝑥(𝑡+𝜏)+𝑃(𝑡+𝜏)𝑡+𝜏𝑃(𝑡+𝜏)^∞_𝑡+𝜏𝑄₁(𝑠)𝑥𝑠−𝜎₁−𝑄₂(𝑠)𝑥𝑠−𝜎₂+1𝑑𝑠𝑃(𝑡+𝜏)_𝑡^𝑡+𝜏₁𝑠𝑄₁(𝑠)𝑥𝑠−𝜎₁−𝑄₂(𝑠)𝑥𝑠−𝜎₂𝑑𝑠+𝐻(𝑡+𝜏)−𝑀𝑃(𝑡+𝜏),𝑡≥𝑡₁(𝑇𝑥)𝑡₁,𝑡₀≤𝑡≤𝑡₁. (2.41)
Clearly, 𝑇𝑥 is continuous. For every 𝑥∈𝐴 and 𝑡≥𝑡₁, using (2.37) and (2.38), we get(𝑇𝑥)(𝑡)≤−1+𝑥(𝑡+𝜏)−1𝑝(𝑡+𝜏)𝑝(𝑡+𝜏)^∞_𝑡1𝑠𝑄₂(𝑠)𝑥𝑠−𝜎₂𝑑𝑠+𝐻(𝑡+𝜏)−𝑀≤𝑝(𝑡+𝜏)1+𝑁₄𝑝₁+𝑁₄𝑝₁^∞_𝑡1𝑠𝑄₂𝑝(𝑠)𝑑𝑠+¹−1𝑝₁≤𝑁₄. (2.42)Furthermore, from (2.36) and (2.38), we have(𝑇𝑥)(𝑡)≥−1+𝑥(𝑡+𝜏)+1𝑝(𝑡+𝜏)𝑝(𝑡+𝜏)^∞_𝑡1𝑠𝑄₁(𝑠)𝑥𝑠−𝜎₁𝑑𝑠+𝐻(𝑡+𝜏)−𝑀≥𝑝(𝑡+𝜏)1+𝑀₄𝑝₂−𝑁₄𝑝₁^∞_𝑡1𝑠𝑄₁𝑝(𝑠)𝑑𝑠−¹−1𝑝₁≥𝑀₄. (2.43)Thus we prove that 𝑇𝐴⊂𝐴. To apply the contraction principle, we need to prove that 𝑇 is a contraction mapping on 𝐴 since 𝐴 is a bounded, closed, and convex subset of 𝑋.
Now, for 𝑥₁,𝑥₂∈𝐴 and 𝑡≥𝑡₁, we have𝑇𝑥₁(𝑡)−𝑇𝑥₂≤1(𝑡)𝑝₁𝑥₁−𝑥₂1+^∞_𝑡1𝑠𝑄₁(𝑠)+𝑄₂(𝑠)𝑑𝑠=𝑞₄𝑥₁−𝑥₂, (2.44)where we used sup norm. From (2.35), we obtain 𝑞₄<1, which completes the proof of Case (iv).
Example 2.5. Consider the second-order neutral delay differential equation𝑥(𝑡)−1+𝑒^−𝑡𝑥(𝑡−1)^′′+𝑒^−𝑡−1𝑥(𝑡−1)−𝑒^−𝑡−1𝑥(𝑡−1)=ℎ(𝑡),𝑡≥1, (2.45)where ℎ(𝑡)=𝑒^−𝑡+4𝑒^−2𝑡+1−𝑒^−𝑡+1. This equation has a nonoscillatory solution 𝑦=𝑒^−𝑡 since the sufficient conditions—in Case (iv) of Theorem 2.1—are satisfied.

Acknowledgments

This work was supported by Natural Science Foundations of Shanxi Province (2007011019), by the special Scientific Research Foundation for the subject of doctor in university (20060110005), and by Program for New Century Excellent Talents in University (NCET050271). The authors are very grateful to the referees for their useful comments.